Mathematics - Probability Question with Solution | TestHub

$P(C/B)=\frac{P(B\cap C)}{P(B)}=\frac{n(B\cap C)}{n(B)}$

The event $(B\cap C)$ is the same as selecting (1W, 1R), (2W, 2R), (3W, 3R), (4W, 4R), (5W, 5R), and at least one but not all black balls when considering all balls different.

$$\begin{array}{rl}& n(B\cap C)=\left({}^6{C}_1{}^6{C}_1+{}^6{C}_2{}^6{C}_2+\dots \dots +{}^6{C}_5\cdot {}^6{C}_5\right)\cdot \left({}^6{C}_1+{}^6{C}_2+\dots \dots +{}^6{C}_5\right)\\ & =\left({}^{12}{C}_6-2\right)\left(2^6-2\right)\\ & \Rightarrow P(C/B)=\frac{{}^{12}{C}_6-2}{{\left(2^6-2\right)}^2}\end{array}$$

(Though the number of different ways for event $B\cap C$ is $5^2$ when balls are identical, i.e., selecting (1W, 1R), (2W, 2R) .....(5W, 5R) → 5 ways and selecting at least one but not all black balls → 5 ways.)